Suppose a group $G$ acts faithfully on a set of five elements, inducing two orbits of size $3$ and $2$ respectively. What group may $G$ be?

There is clearly a homomorphism $G \mapsto S_3$ and another $G \mapsto S_2$.

$|G|$ cannot be greater than $|S_2 \times S_3| = 12$, the total number of simultaneous permutations, or its action would not be faithful.

It seems to me that $D_3=S_3$, $C_6$, and $S_3 \times C_2$ are possibilities. Is this list correct and exhaustive, and how do I prove it?


2 Answers 2


Put together the following facts: orbit-stabilizer, Lagrange's theorem, $G$ is $\cong$ a subgroup of the symmetry group $S_2\times S_3$ by faithfulness, and classification of groups of order $6$.


$G$ is a subgroup of $S_3 \times S_2$ since $G$ acts faithfully and permutes the orbits

Due to the counting formula, the order of the orbits divide the order of the group: $|G|$ is divisible by both $3$ and $2$, so it is a multiple of $6$.

So far, $G$ could only be $D_3$, $C_6$ or $S_3 \times S_2$. We have yet to show each of these is possible:

  • $G = D_3$ is possible since it is the group of symmetries of a triangular bipyramid, whose $5$ vertices satisfy the hypotheses.

  • $G = C_6$ with generator $x$ satisfies the hypotheses for $G$ if $x$ induces any nontrivial permutation on both sets.

  • $G = S_3 \times S_2$ satisfies the hypotheses with its elements $(\sigma,\tau)$ applying $\sigma$ to the three element orbit and $\tau$ to the two element orbit.

Therefore, $D_3$, $C_6$, and $S_3 \times S_2$ are all possibilities.


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